Here, a = 5, d = - 3
\(\because\)  l = a + (n -1)d
\(\because\)- 49 = 5 + (n -1)(- 3)
\(\Rightarrow\)-49 = 5-3n + 3
\(\Rightarrow\) 3n = 49 + 5 + 3
\(\Rightarrow\)n = 57/3 = 19^th term.

a_n=3+4n
a₁=3+4x1=7
a₂=3+4x2=11
a₃=3+4x3=15
Sequence is, 7,11,15

a = 10,
d = 10.5 - 10 = 0.5
a₁₀ = a + 9d = 10 + 9 x 0.5 = 14.5

Here, we have
a₂ -a₁ = \(\sqrt{8}\) - \(\sqrt{2}\) = \(2\sqrt{2}\) - \(\sqrt{2}\) = \(\sqrt{2}\)
 a₃ - a₂ = \(\sqrt{18}\) - \(\sqrt{8}\) =  \(3\sqrt{2}\) - \(2\sqrt{2}\) = \(\sqrt{2}\)
a₄ - a₃ = \(\sqrt{32}\)-\(\sqrt{18}\)=4\(\sqrt{2}\) -3\(\sqrt{2}\) = \(\sqrt{2}\) and so on. Since, the difference of any two consecutive terms is same. Therefore, the given list of numbers forms an AP and its common difference (d) is \(\sqrt{2}\)
Now, next three terms of this AP are,
a₅ = a₄ + d = 4\(\sqrt{2}\)+ \(\sqrt{2}\) = 5\(\sqrt{2}\)
a₆ = a₅ + d = 5\(\sqrt{2}\) + \(\sqrt{2}\) = 6\(\sqrt{2}\)
and a₇ = a₆ + d = 6 \(\sqrt{2}\) + \(\sqrt{2}\) = 7 \(\sqrt{2}\)

Here, we have   
\(a_{2}-a_{1}=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}\),
\(a_{3}-a_{2}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\)
\(a_{4}-a_{3}=\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\) and so on.
Since, the difference of any two consecutive terms is same. therefore, the given list of numbers forms an AP and its common difference (d) is \(\frac{1}{2}\).
Now, next three terms of this AP are,
a₅ = a₄ + d = \(\frac{7}{2}+\frac{1}{2}\)
[\(\because\) a₅ = a + 4d = a + 3d + d = a₄ + a₅]
\(=\frac{7+1}{2}=\frac{8}{2}=4\)
\(a_{6}=a_{5}+d=4+\frac{1}{2}=\frac{9}{2}\)
and \(a_{7}=a_{6}+d=\frac{9}{2}+\frac{1}{2}=\frac{9+1}{2}=\frac{10}{2}=5\)

Given, a = -2, d = 0
The first four terms of the AP are -2, -2, -2, and -2.

111222333

13, 17, 21

Let a be the first term and d be the common difference of the AP.
Gien,         T₂ = \(7\frac{3}{4}\)
\(\Rightarrow\)    a + d = \(\frac{31}{4}\)           ...(i)
and         T₃₁ = \(\frac{1}{2}\)
\(\Rightarrow\)    a + 30d = \(\frac{1}{2 }\)
Subtracting (i) from (ii), we get
29d = \(\frac{1}{2}-\frac{31}{4}=-\frac{29}{4}\)   \(\Rightarrow\) d = - \(\frac{1}{4}\)
Putting the value of d in (i), we get
\(a-\frac{1}{4}=\frac{31}{4}\)      \(\Rightarrow\)  \(a=\frac{31}{4}+\frac{1}{4}=\frac{32}{4}=8\)
Let the number of terms be n, so that
T_n = \(-\frac{13}{2}\)
i.e.,  a + ( n - 1 )d = - \(\frac{13}{2}\)  \(\Rightarrow\)   \(8-\frac{n}{4}+\frac{1}{4}=-\frac{13}{2}\)
\(\Rightarrow\)   32 - n + 1 = - 26 \(\Rightarrow\) n = 59
Hence, first term = 8 and number of terms = 59.

Numbers divisible by 9 between 100 and 200 are 108, 117, 126, ..., 198.
Here, a = 108,  d = 9, a_n = 198
  an = a + ( n - 1 )d    \(\Rightarrow\)      198 = 108 + ( n - 1 )9 
\(\Rightarrow\)    198 = 108 + 9n - 9    \(\Rightarrow\)  198 = 99 +  9n
\(\Rightarrow\)    198 - 99 = 9n        \(\Rightarrow\)  \({99\over9}\) = n      \(\Rightarrow\) 11 = n
and  S_n = \({n\over 2}[2a+(n-1)d]\)
S₁₁ = \({11\over2}\) [ 2 X 108 + ( 11 - 1 )9]
= \({11\over2}\) [ 216 + 90 ] = \({11\over2}\times306\) = 1683